Basics of Fluid Mechanics, 2014a

112 CHAPTER 4. FLUIDS STATICS
F x = ρgbr2
2
(
1 − cos 2 (θ 0 ) )
Alternative way to do this calculation is by calculating
the pressure at mid point and then
multiply it by the projected area, A x (see Figure
4.29) as
F x = ρg
x c
A { }}
x
{ }} {
{ r sin θ 0
brsin θ 0 = ρgbr
2 2
sin 2 θ
A △ = r 2 sin θ cos θ
r
A arc = θr2
2
Fig. -4.29. Area above the dam arc subtract
triangle.
Notice that dA x (cos θ) and A x (sin θ) are different,
why?
The values to evaluate the last equation are provided in the question and simplify
subsidize into it as
F x =
1000 × 9.8 × 4 × 2
2
sin(45 ◦ ) = 19600.0[N]
Since the last two equations are identical (use the sinuous theorem to prove it
sin 2 θ + cos 2 =1), clearly the discussion earlier was right (not a good proof LOL 14 ).
The force in the y direction is the area times width.
V
{ ⎛ }} ⎞ {
A
{ }} {
θ 0 r 2
F y = − ⎜ − r2 sin θ 0 cos θ 0
⎟ bgρ∼ 22375.216[N]
⎝ 2 2 ⎠
The center area ( purple area in Figure 4.29) should be calculated as
y c = y c A arc − y c A triangle
A
The center area above the dam requires to know the center area of the arc and triangle
shapes. Some mathematics are required because the shift in the arc orientation. The
arc center (see Figure 4.30) isat
y carc = 4 r ( )
sin2 θ
2
3 θ
14 Well, it is just a demonstration!